Let \(C\) be a curve of genus \(g\) defined over the finite field \(\FF_q\), and let \(N\) be the number of points of \(C\) rational over \(\FF_q\). Then \(N=1-a+q\), with \(|a| \leqslant 2 g \sqrt{q}\).
To prove this, we consider \(C\) as a curve over the algebraic closure \(k\) of \(\FF_q\). Let \(f: C \rightarrow C\) be the \(k\)-linear Frobenius morphism obtained by taking \(q\) th powers, which makes sense since \(C\) is defined over \(\FF_q\), so \(X_q \cong X\) (See \(V, 2.4.1\)).
Let \(\Gamma \subseteq C \times C\) be the graph of \(f\), and let \(\Delta \subseteq C \times C\) be the diagonal.
Show that \(\Gamma^2=q(2-2 g)\), and \(\Gamma . \Delta=N\). Then apply (Ex. 1.9) to \(D=r \Gamma+s \Delta\) for all \(r\) and \(s\) to obtain the result.
See (App. C, Ex. 5.7) for another interpretation of this result.
Solution
Let \[ F=(f,\id_C):C\times C\longrightarrow C\times C. \]
1. The \(q\)-power Frobenius \(f:C\to C\) is finite of degree \(q\).
Write \(q=p^r\). The \(k\)-linear \(p\)-power Frobenius on a smooth curve over the perfect field \(k\) has degree \(p\); this is recorded in D-IV2FROBTWIST. The \(q\)-power Frobenius is its \(r\)-fold iterate, with the intervening Frobenius twists identified with \(C\) because the curve is defined over \(\FF_q\). Multiplicativity of degree under composition therefore gives \[ \deg f=p^r=q. \]
2. Scheme-theoretically,\[ \Gamma=F^*\Delta, \]and\[ \boxed{\Gamma^2=q(2-2g)}. \]
The inverse image of the diagonal under \(F\) consists of pairs \((P,Q)\) with \[ f(P)=Q, \] which is exactly the graph of \(f\); the same equality holds scheme-theoretically because both are defined by pulling back the diagonal ideal.
1. The projection formulafor intersection numbers therefore gives\[ (F^*\Delta)^2 = q\Delta^2. \]Exercise V.1.6, proved on P-AGH516DIAGONAL, gives\[ \Delta^2=2-2g. \]Substitution yields\[ \Gamma^2=q(2-2g). \]
3. The closed points of \(\Gamma\cap\Delta\) are exactly the\(\FF_q\)-rational points of \(C\).
A geometric point \((P,P)\) of the diagonal lies on the graph exactly when \[ f(P)=P. \] The fixed geometric points of the \(q\)-power Frobenius are precisely the points defined over \(\FF_q\). Thus the underlying set of \(\Gamma\cap\Delta\) is in bijection with \(C(\FF_q)\).
4. Every point of \(\Gamma\cap\Delta\) has intersection multiplicity one, so\[ \boxed{\Gamma\cdot\Delta=N}. \]
The differential of Frobenius is zero. At a fixed point \(P\), under \[ T_{(P,P)}(C\times C)=T_PC\oplus T_PC, \] the tangent line to the graph is \[ T_{(P,P)}\Gamma=\{(v,0):v\in T_PC\}, \] while the tangent line to the diagonal is \[ T_{(P,P)}\Delta=\{(v,v):v\in T_PC\}. \] These two one-dimensional subspaces meet only at zero and span the tangent space of the surface. Hence the two smooth curves meet transversally at every fixed point, so every local intersection multiplicity is one. Step 3 now gives \[ \Gamma\cdot\Delta=\#C(\FF_q)=N. \]
5. With\[ l=C\times\{Q\}, \qquad m=\{Q\}\times C \]for a geometric point \(Q\in C\), the divisors \(\Gamma\) and \(\Delta\) have types\[ \Gamma:(q,1), \qquad \Delta:(1,1) \]in the notation of Exercise V.1.9.
1. Hence\[ \Gamma\cdot l=q. \]The diagonal meets each ruling transversally in one point, so\[ \Delta\cdot l=\Delta\cdot m=1. \]This gives the stated types.
6. For integers \(r,s\), put\[ D=r\Gamma+s\Delta. \]Then\[ D^2 =(2-2g)(qr^2+s^2)+2rsN \]and \(D\) has type\[ (rq+s,r+s). \]
The square formula follows from steps 2 and 4 together with \(\Delta^2=2-2g\): \[ \begin{aligned} D^2 &=r^2\Gamma^2+2rs(\Gamma\cdot\Delta)+s^2\Delta^2\\ &=(2-2g)(qr^2+s^2)+2rsN. \end{aligned} \] The type is additive, so step 5 gives \[ D\cdot l=rq+s, \qquad D\cdot m=r+s. \]
7. For every pair of integers \(r,s\),\[ rs(N-q-1)\leq g(qr^2+s^2). \]
Apply the Castelnuovo–Severi inequality from Exercise V.1.9, P-AGH519HODGEINDEX, to the divisor \(D\) of step 6: \[ D^2\leq2(D\cdot l)(D\cdot m). \] Substituting the formulas from step 6 gives \[ (2-2g)(qr^2+s^2)+2rsN \leq 2(rq+s)(r+s). \] Dividing by two and collecting terms yields \[ rs(N-q-1)\leq g(qr^2+s^2). \]
8. One has\[ N-q-1\leq2g\sqrt q. \]
Take \(r,s>0\) in step 7 and divide by \(rs\): \[ N-q-1 \leq g\left(q\frac rs+\frac sr\right). \] As \(r/s\) ranges over the positive rational numbers, these ratios are dense in \(\RR_{>0}\). For \(x>0\), the arithmetic–geometric mean inequality gives \[ qx+\frac1x\geq2\sqrt q, \] with equality at \(x=1/\sqrt q\). Hence the infimum over positive rational \(x\) is \(2\sqrt q\). Taking the infimum in the displayed family of upper bounds gives \[ N-q-1\leq2g\sqrt q. \]
9. One has\[ N-q-1\geq-2g\sqrt q. \]
7. Dividing by the negative number \(rs\)reverses the inequality:\[ N-q-1 \geq g\left(q\frac rs+\frac sr\right). \]Writing \(x=r/s<0\) and \(t=-x>0\) gives\[ qx+\frac1x = -\left(qt+\frac1t\right) \leq -2\sqrt q. \]The negative rational numbers are dense in \(\RR_{<0}\), so the supremum of theright-hand side over such rational \(x\) is \(-2g\sqrt q\). Taking this supremumgives the asserted lower bound.
10. Setting\[ a=q+1-N, \]one has\[ \boxed{N=1-a+q} \qquad\text{and}\qquad \boxed{|a|\leq2g\sqrt q}. \]
Steps <1>8–<1>9 give \[ |N-q-1|\leq2g\sqrt q. \] Since \(a=-(N-q-1)\), the same bound holds for \(|a|\), and the defining equation for \(a\) rearranges to \(N=1-a+q\).
11. Q.E.D.
Steps <1>1–<1>4 prove the two requested intersection identities, and steps <1>5–<1>10 apply Exercise V.1.9 to obtain the stated Riemann-hypothesis bound.